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3 years ago

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∆ABC with Prove: Statement Reason 1. given 2. ∠CAB ≅ ∠EDB ∠ACB ≅ ∠DEB If two parallel lines are cut by a transversal, the corres

ponding angles are congruent. 3. ∆ABC ~ ∆DBE AA criterion for similarity 4. Corresponding sides of similar triangles are proportional. 5. AB = AD + DB CB = CE + EB segment addition 6. Substitution Property of Equality 7. division 8. Subtraction Property of Equality What is the missing statement in the proof? Scroll down to see the entire proof. A. B. C. D.

1 answer:

avanturin [10]3 years ago

5 0

Answer with Step-by-step explanation:

We are given that a triangle ABC

$DE\parallel AC$

We prove that $\frac{AD}{DB}=\frac{CE}{EB}$

We have to find the missing step in given proof.

Proof:

1.Statement: $DE\parallel AC$

Reason: Given

2.Statement: $\angle CAB\cong \angle EDB. \angle ACB\cong\angle DEB$

Reason:If two parallel lines are cut by a transversal line then, congruent angles are congruent.

3.Statement: $\triangle ABC\cong \triangle DBE$

Reason: AA criterion for similarity

4.Statement: $\frac{AB}{DB}=\frac{CB}{EB}$

Reason:Corresponding sides of similar triangles are proportional

5. $AB=AD+DB, CB=CE+EB$

Reason: Segment addition property

6.Statement: $\frac{AD+DB}{DB}=\frac{CE+EB}{EB}$

Reason:Substitution property of equality

7.Statement: $\frac{AD}{DB}+1=\frac{CE}{EB}+1$

Reason:Division property

8.Statement: $\frac{AD}{DB}=\frac{CE}{EB}$

Reason: Subtraction property of equality

Hence, proved.

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The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

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d. Find the Variance of X

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